Question: Simplify the following expression: $y = \dfrac{9x^2- 49x+20}{x - 5}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(9)}{(20)} &=& 180 \\ {a} + {b} &=& &=& {-49} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $180$ and add them together. The factors that add up to ${-49}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-4}$ and ${b}$ is ${-45}$ $ \begin{eqnarray} {ab} &=& ({-4})({-45}) &=& 180 \\ {a} + {b} &=& {-4} + {-45} &=& -49 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({9}x^2 {-4}x) + ({-45}x +{20}) $ Factor out the common factors: $ x(9x - 4) - 5(9x - 4)$ Now factor out $(9x - 4)$ $ (9x - 4)(x - 5)$ The original expression can therefore be written: $ \dfrac{(9x - 4)(x - 5)}{x - 5}$ We are dividing by $x - 5$ , so $x - 5 \neq 0$ Therefore, $x \neq 5$ This leaves us with $9x - 4; x \neq 5$.